Conditional Statements (IF, Else) not working properly in PHP

Question

I am trying to apply some if else statements in my code but getting a bit confused. I want a result if I got value=0 answer should be One. if I got value=1 answer should be Two and if I got any value greater then 1 value should be Three. Following is my code: -

$value = 1
if($value < 1)
{
  echo "One";
}
else if($value === '1')
{
  echo "Two";
}
else if($value >= 2)
{
  echo "Three";
}

The problem is that it is not giving me results if value is = 1.

`1 == '1'`, but `1 !== '1'` (they are of different types). See http://php.net/manual/en/language.operators.comparison.php — cmbuckley, Dec 17 '18 at 16:34
`=== '1'` does not equal to `=== 1` so just change it if your `$value` is integer — Alex, Dec 17 '18 at 16:34
`$value = 1` < did you even close it off? It's missing a semi-colon here. — Funk Forty Niner, Dec 17 '18 at 16:35

score 1 · Answer 1 · answered Dec 17 '18 at 16:34

1

change this else if($value === '1') to else if($value == 1)

answered Dec 17 '18 at 16:34

Giorgi Lagidze

773
4
24

still not working – Imran Dec 17 '18 at 16:40

score 0 · Answer 2 · answered Dec 17 '18 at 16:36

=== matched both value and type that's why 1==='1' is false, See Ref.

$a === $b Identical TRUE if $a is equal to $b, and they are of the same type.

<?php
$value = 1;
if($value < 1)
{
    echo "One";
}
else if($value == '1')
{
    echo "Two";
}
else if($value >= 2)
{
    echo "Three";
}
?>

DEMO: https://3v4l.org/avOWc

Conditional Statements (IF, Else) not working properly in PHP

2 Answers2